Question

Answer 1

这是一个非常基本的示例，没有错误处理，并且使用非常严格定义的XML文件，但您应该将其作为开头并继续使用以下链接：

代码：

import os
import os.path
from xml.dom.minidom import parse


def parse_file(path):
    files = []
    try:
        dom = parse(path)
        for filetag in dom.getElementsByTagName('File'):
            type = filetag.getElementsByTagName('Type')[0].firstChild.data
            if type == 'config':
                path = tag.getElementsByTagName('Path')[0].firstChild.data
                files.append(path)
        dom.unlink()
    except:
        raise
    return files


def main():
    files = []
    for root, dirs, files in os.walk('/var/packs'):
        if 'info.xml' in files:
            files += parse_file(os.path.join(root, 'info.xml'))
    print 'The list of desired files:', files


if __name__ == '__main__':
    main()

Answer 2

使用lxml.etree和XPath：

files = []
for root, dirnames, filenames in os.walk('/var/packs'):
    for filename in filenames:
        if filename != 'info.xml':
            continue
        tree = lxml.etree.parse(os.path.join(root, filename))
        files.extend(tree.getroot().xpath('//File[Type[text()="config"]]/Path/text()'))

如果lxml不可用，您也可以使用标准库中的etree API：

files = []
for root, dirnames, filenames in os.walk('/var/packs'):
    for filename in filenames:
        if filename != 'info.xml':
            continue
        tree = xml.etree.ElementTree.parse(os.path.join(root, filename))
        for file_node in tree.findall('File'):
            type_node = file_node.find('Type')
            if type_node is not None and type_node.text == 'config':
                path_node = file_node.find('Path')
                if path_node is not None:
                    files.append(path_node.text)

Answer 3

把它写在我的头顶，但是这里。我们将使用os.path.walk以递归方式下降到您的目录和minidom中进行解析。

import os
from xml.dom import minidom

# opens a given info.xml file and prints out "Path"'s contents
def parseInfoXML(filename):
    doc = minidom.parse(filename)
    for fileNode in doc.getElementsByTagName("File"):
        # warning: we assume the existence of a Path node, and that it contains a Text node
        print fileNode.getElementsByTagName("Path")[0].childNodes[0].data
    doc.unlink()

def checkDirForInfoXML(arg, dirname, names):
    if "info.xml" in names:
        parseInfoXML(os.path.join(dirname, "info.xml"))

# recursively walk the directory tree, calling our visitor function to check for info.xml in each dir
# this will include packs as well, so be sure that there's no info.xml in there
os.path.walk("/var/packs" , checkDirForInfoXML, None)

不是最有效的方法，我敢肯定，但如果你不指望任何错误/无论如何都会这样做。

解析xml文件并创建文件列表

3 个答案: